#dy/dx of logx/√(x^2-a^2)
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can u show ur solving once
alright
$\frac{dy}{dx}=\frac{\frac{d logx}{dx}\cdot\sqrt{x^{2}-a^{2}} - logx \cdot \frac{d \sqrt{x^{2}-a^{2}}}{dx}}{x^{2}-a^{2}}$
what
uh wait
also
derivate of log x is not 1/x
its $\frac{1}{x lna}$ where a is the base
rev #tea4life
i havent derivated it
yet
i kept it for the quoatient rule
isnt logx just log base e and x
rev #tea4life
yo i converted this and
what happened to the x in x/√(x^2-a^2)
okay
e
then it's natural log
yea
i thought it was base 10
1/x
yea
no
rev #tea4life
I'd just use quotient rule
v=√(x^2-a^2)
lemme try with quotient rule
so that's (u'v-uv')/v²
the answer is weird too
its correct
Hello
simplifying what i get
yea
here the sign between the 2 frac is supposed to be -ve, not +ve
in the 3rd step
silly error mb
yeah sure
ur welcome
+close